Respuesta :
Answer:
a)Blue car speed in t =2.6 s : v=11.8m/s
b)Blue car speed in t =7.2 s Β : v=14.19m/s
c)Blue car distance : d=226.31m
d)acceleration of the blue car applying the brakes a= 3.6 m/sΒ²
e)total time for the blue car: t=21.54s
f) acceleration of the yellow car: a=1.09 m/sΒ²
Explanation:
Blue car kinematics:
Movement 1) time interval:(0,3.3)s: uniformly accelerated movement a=4.3m/sΒ², vβ=0
dβββ= vβ*t - Β½ a*tΒ²=0(3.3)*+Β½* 4.3*3.3Β²=23.41m
vβ=vβ+a*t=o+4.3*3.3
vβ=14.19m/s
Movement 2) time interval:(3.3,14.3)s Uniform movement
V is constant=vβ=14.19m/s
dβββ=v*t=14.19*14.3=202.9m
Movement 3) applying the brakes : final speed=0
vβ=14.19m/s Β ,vβ=0
dβββ=253.26-22.41-202.9= 27.95m
vβΒ²=vββΒ²-2*a*dβββ
0=14.19Β²-2*a*27.95
a=14.19Β²Γ·2*27.95
a=3.6m/sΒ²
vβ=vβ-a*tβββ
0=14.19-3.6t
tβββ=14.19Γ·3.6
tβββ=3.94s
yellow car kinematics:
dβββ= vβ*t +Β½ a*tΒ²
dβββ=dβββ+dβββ+dβββ=23.41m+202.9m+27.95m=254.26m
t=3.3+14.3+3.94=21.54 s
Problem development
a)Blue car speed in t =2.6 s Β : t(0,3.3)s: Movement 1)
vβββ=vβ+a*t=0+4.3*2.6
v(2.6s)=0+4.3*2.6
v(2.6s)=11.18m/s
b)Blue car speed in t =7.2 s Β :t(3.3,14.3)s, Movement 2) : v is constant
v(7.2s)=14.19m/s
c)Blue car distance : dβββ=Movement 1+Movement 2
dβββ=dβββ+dβββ=23.41+202.9=226.31m
d)acceleration of the blue car applying the brakes :Movement 3)
a=3.6m/sΒ²
e)total time for the blue car:
time(Movement 1)+time(Movement 2)+time(Movement 3)
tβββ=3.3+14.3+3.94=21.54 s
f) acceleration of the yellow car
vβ=0
dβββ=254.26m
tβββ=21.54
dβββ= vβ*t + Β½ a*tΒ²
254.26=0+ Β½ *a*21.54Β²
a= 2*254.26 Γ· 21.54Β²
a=1.09 m/sΒ²